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means_._note [2015/06/20 17:04] nikolaj |
means_._note [2015/06/20 17:05] nikolaj |
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| | @#FFBB00: definiendum | @#FFBB00: $\langle f\rangle:=I(f\cdot w)\cdot I(w)^{-1}$ | | | @#FFBB00: definiendum | @#FFBB00: $\langle f\rangle:=I(f\cdot w)\cdot I(w)^{-1}$ | | ||
| - | e.g. $\langle f\rangle_{[a,b]}:=\dfrac{\int_a^bf(x)\,{\mathrm dx}}{b-a}$ | + | == Real functions == |
| + | |||
| + | E.g. $\langle f\rangle_{[a,b]}:=\dfrac{\int_a^bf(x)\,{\mathrm dx}}{b-a}$ | ||
| where $[a,b]\subseteq{\mathbb R}$ and $w(x):=1$. | where $[a,b]\subseteq{\mathbb R}$ and $w(x):=1$. | ||
| - | == Note == | + | == Minus twelve == |
| I use this in the context of [[Minus twelve . Note]]. For $z\in(0,1)$, we find | I use this in the context of [[Minus twelve . Note]]. For $z\in(0,1)$, we find | ||
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| $\sum_{k=0}^\infty \langle q\mapsto q\,z^q\rangle_{[k,k+1]}=\dfrac{1}{\ln(z)^2}$, | $\sum_{k=0}^\infty \langle q\mapsto q\,z^q\rangle_{[k,k+1]}=\dfrac{1}{\ln(z)^2}$, | ||
| - | i.e. | + | i.e. (see [[Natural logarithm of complex numbers]]) |
| $\sum_{k=0}^\infty \left(k\,z^k-\langle q\mapsto q\,z^q\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$ | $\sum_{k=0}^\infty \left(k\,z^k-\langle q\mapsto q\,z^q\rangle_{[k,k+1]}\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$ | ||
| - | |||
| - | See also [[Natural logarithm of complex numbers]]. | ||
| ----- | ----- | ||
| === Requirements === | === Requirements === | ||
| [[Function integral]] | [[Function integral]] | ||
